\(\sqrt{281108+291108\sqrt{1+x}}=1+\sqrt{281108-291108\sqrt{1+x}}\)
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DN
10 tháng 10 2017
\(\sqrt{281108+291108\sqrt{1+x}}-\sqrt{281108+291108\sqrt{1+x}}=1\)=1
điều này vô lí
vì hiệu cùa hai số bằng nhau phải bằng 0
=> phương trình vô nghiệm
NN
28 tháng 10 2020
\(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
28 tháng 10 2020
Để xem nào ...
Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
ĐKXĐ : x > 0 ; x khác 1
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
22 tháng 2 2017
GIAO LUU
\(Lim_{x\rightarrow vc}=\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \)
\(\Leftrightarrow Lim_{x\rightarrow vc}=\frac{\sqrt{\frac{x+\sqrt{x}}{x}}}{\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}+1}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{x+\sqrt{x}}{x^2}}}+1}\\ \)
\(\Leftrightarrow\frac{Lim}{x\rightarrow+vc}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x^3}}}}+1}=\frac{\sqrt{1+\frac{1}{+vc}}}{\sqrt{1+\sqrt{\frac{1}{+vc}+\frac{1}{+vc}}}+1}=\frac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}}+1}=\frac{1}{2}\)
27 tháng 8 2023
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
27 tháng 8 2023
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(B=\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{x+2\sqrt{x}+\sqrt{x}+1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{2x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(B=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
\(\sqrt{281108+291108\sqrt{1+x}}=1+\sqrt{281108-291108\sqrt{1+x}}\)
\(\Leftrightarrow\sqrt{281108+291108\sqrt{1+x}}-\sqrt{281108-291108\sqrt{1+x}}=1\)
\(\Leftrightarrow\left(\sqrt{281108+291108\sqrt{1+x}}-\sqrt{281108-291108\sqrt{1+x}}\right)^2=1\)
\(\Leftrightarrow281108+291108\sqrt{1+x}+281108-291108\sqrt{1+x}-2\sqrt{\left(281108+291108\sqrt{1+x}\right)\left(281108-291108\sqrt{1+x}\right)}=1\)
\(\Leftrightarrow562216-2\sqrt{\left(281108+291108\sqrt{1+x}\right)\left(281108-291108\sqrt{1+x}\right)}=1\)
\(\Leftrightarrow562215-2\sqrt{\left(281108+291108\sqrt{1+x}\right)\left(281108-291108\sqrt{1+x}\right)}=0\)
P/s : Đến đây tự giải nha số to quá
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